- #1
Robin04
- 260
- 16
Homework Statement
Solve the following vector equation for ##\vec{y}##. ##\vec{a}##, and ##\vec{b}## are linearly independent vectors of the three dimensional space.
##\vec{a} \times (8\vec{y}+\vec{b}) = \vec{b}\times(-5\vec{y}+\vec{a})##
Homework Equations
The Attempt at a Solution
First I developed the brackets
##\vec{a}\times8\vec{y}+\vec{a}\times\vec{b} = \vec{b}\times(-5\vec{y})+\vec{b}\times\vec{a}##
I subtracted ##\vec{b}\times(-5\vec{y})## and ##\vec{a}\times\vec{b}## from both sides, so I get
##\vec{a}\times8\vec{y}-\vec{b}\times(-5\vec{y})=\vec{b}\times\vec{a}-\vec{a}\times\vec{b}## which also equals to
##\vec{y}\times(-8\vec{a}-5\vec{b}) = 2\vec{b}\times\vec{a}##
Then I expressed ##\vec{y}## as the linear combination of ##\vec{a}##, ##\vec{b}## and ##\vec{b}\times\vec{a}##
##\vec{y} = \alpha\vec{a}+\beta\vec{b}+\gamma\vec{b}\times\vec{a}## then I multiplied it (cross product) with ##-8\vec{a}-5\vec{b}##
##\vec{y}\times(-8\vec{a}-5\vec{b}) = -8\alpha(\vec{a}\times\vec{a})-5\alpha(\vec{a}\times\vec{b})-8\beta(\vec{b}\times\vec{a})-5\beta(\vec{b}\times\vec{b}) + \gamma(\vec{b}\times\vec{a})\times(-8\vec{a}-5\vec{b})##
##=(5\alpha-8\beta)(\vec{b}\times\vec{a})-8\gamma[\vec{a}(\vec{b}\vec{a})-\vec{b}(\vec{a}\vec{a})]-5\gamma[\vec{a}(\vec{b}\vec{b})-\vec{b}(\vec{a}\vec{b})]##
From the equation this has to be equal to ##2\vec{b}\times\vec{a}## so ##5\alpha-8\beta = 2## and ##\gamma = 0##
But if I make up an arbitrary ##\vec{y}## vector of this form and choose random ##\vec{a}## and ##\vec{b}## (linearly independent of course) it doesn't satisfy the equation. Where did I mess up?